Finding Segment Lengths with Algebra | Geometry | LevelKite

A segment is the part of a line made of two points, called endpoints, and all the points between them. You write $\overline{AB}$ for the segment and $AB$ for its length, the distance between its endpoints. A segment's length is always positive.

Sometimes you know a length only as an expression, such as $AB = 2x + 3$ . Think of it as a length waiting for a value of $x$ . To find that value, you need a fact that ties the lengths together. The fact gives you an equation, and solving the equation turns each expression into a number. You already know the Segment Addition Postulate and how to solve multi-step equations, so now you'll use them together.

Key concept

A given fact gives you the equation

To find a length with algebra, use only facts that are stated in the text or shown by a mark. No figure in this lesson is drawn to scale, so nothing comes from how a figure looks.

Write an equation from a given fact. For example, suppose the text states that $B$ is between $A$ and $C$ , meaning $B$ lies on $\overline{AC}$ but is not an endpoint. Then the Segment Addition Postulate gives the equation below.
Solve the equation, then substitute the value of $x$ to find each length.
Check each length. Reject a value of $x$ that makes any length zero or negative.

\text{If } B \text{ is between } A \text{ and } C, \text{ then } AB + BC = AC.

Not drawn to scale. The figure marks the three lengths from the example below:

AB = 2x + 3

,

BC = 4x - 1

, and

AC = 20

.

Worked example

Two parts and a known whole

Given: point $B$ is between $A$ and $C$ , $AB = 2x + 3$ , $BC = 4x - 1$ , and $AC = 20$ . The figure marks these three lengths. Find: $x$ , $AB$ , and $BC$ .

Point B is stated in the text to be between A and C. That is the "if" part of the Segment Addition Postulate, so its "then" part holds: AB + BC = AC. Replace each length with its given value (Substitution Property of Equality).
$AB + BC = AC, \text{ so } (2x + 3) + (4x - 1) = 20$
Combine like terms on the left side: 2x with 4x, and 3 with -1. This changes the form of the left side but not its value.
$6x + 2 = 20$
Subtract 2 from both sides (Subtraction Property of Equality), which leaves the term with x alone.
$6x = 18$
Divide both sides by 6 (Division Property of Equality). The value x = 3 is not a length yet. It is the number that turns each expression into a length.
$x = 3$
Substitute x = 3 into each expression to find the lengths (Substitution Property of Equality).
$AB = 2(3) + 3 = 9, \quad BC = 4(3) - 1 = 11$
Check the result. Both lengths are positive, so you keep x = 3. They also add to the given whole, so the values fit the Segment Addition Postulate.
$9 + 11 = 20$

Answer

The value of $x$ is $3$ , so $AB = 9$ and $BC = 11$ .

Tick marks give you another way to write an equation. Congruent segments are segments with equal lengths, and you write $\overline{DE} \cong \overline{EF}$ to say so. When two segments carry the same number of tick marks, the figure is stating, as a given, that they are congruent. Then the definition of congruent segments tells you their lengths are equal.

In the next figure, $\overline{DE}$ is drawn shorter than $\overline{EF}$ , yet each has one tick, so $DE = EF$ . Go by the mark, not the drawing, because the figure is not drawn to scale.

Not drawn to scale. One tick on each of

\overline{DE}

and

\overline{EF}

states that

DE = EF

. The example gives

DE = 3x - 4

and

EF = x + 6

.

Worked example

Equal lengths from tick marks

Given: point $E$ is between $D$ and $F$ , $\overline{DE} \cong \overline{EF}$ as shown by the tick marks in the figure, $DE = 3x - 4$ , and $EF = x + 6$ . Find: $DF$ .

Segments DE and EF each carry one tick mark, so they are given as congruent. By the definition of congruent segments, their lengths are equal. Substitute the given expressions.
$DE = EF, \text{ so } 3x - 4 = x + 6$
Subtract x from both sides (Subtraction Property of Equality).
$2x - 4 = 6$
Add 4 to both sides (Addition Property of Equality).
$2x = 10$
Divide both sides by 2 (Division Property of Equality).
$x = 5$
Substitute x = 5 into each expression. Both lengths are positive, so keep x = 5.
$DE = 3(5) - 4 = 11, \quad EF = 5 + 6 = 11$
Point E is stated to be between D and F, so the Segment Addition Postulate gives the whole length, which is also positive.
$DF = DE + EF = 11 + 11 = 22$

Answer

The length $DF$ is $22$ , with $DE = EF = 11$ .

Solving the equation doesn't always finish the job. An equation can have a solution that no segment could match, because a segment's length must be positive. So after you solve, test each length, not $x$ itself. A negative $x$ is fine when every length it produces is positive, and you must reject any value of $x$ that makes even one length zero or negative. The next example shows a value that fails.

Worked example

A value the check rejects

Given: point $Q$ is between $P$ and $R$ , $PQ = 3x + 8$ , $QR = x - 2$ , and $PR = 2x + 4$ . Find: each length, or show that no segment can have these lengths.

Q is stated to be between P and R, so apply the Segment Addition Postulate and substitute the given expressions.
$(3x + 8) + (x - 2) = 2x + 4$
Combine like terms on the left side.
$4x + 6 = 2x + 4$
Subtract 2x from both sides (Subtraction Property of Equality).
$2x + 6 = 4$
Subtract 6 from both sides (Subtraction Property of Equality).
$2x = -2$
Divide both sides by 2 (Division Property of Equality).
$x = -1$
Substitute x = -1 into each expression and check. The equation holds, since 5 + (-3) = 2, but QR = -3, and a segment's length must be positive. Reject x = -1. The trouble is the negative length, not the negative x.
$PQ = 3(-1) + 8 = 5, \quad QR = -1 - 2 = -3, \quad PR = 2(-1) + 4 = 2$

Answer

The equation's only solution, $x = -1$ , makes $QR = -3$ , so you reject it, and no segment can have these lengths.

Takeaway: A fact that is stated or marked, never the look of a drawing, gives you the equation. Solve it, substitute to find each length, and keep a value of $x$ only when every length it produces is positive.